MOOCULUS微积分-2: 数列与级数学习笔记 2. Series

此课程(MOOCULUS-2 "Sequences and Series")由Ohio State University于2014年在Coursera平台讲授。

PDF格式教材下载 Sequences and Series

本系列学习笔记PDF下载（Academia.edu） MOOCULUS-2 Solution

Summary

• Suppose $(a_n)$ is a sequence with associated series $$\sum_{k=1}^\infty a_k$$ The sequence of partial sums associated to these objects is the sequence $$s_n = \sum_{k=1}^n a_k$$
• Consider the series $$\sum_{k=1}^\infty a_k$$ This series converges if the sequence of partial sums $$s_n = \sum_{k=1}^n a_k$$ converges. More precisely, if $$\lim_{n \to \infty} s_n = L$$ we then write $$\sum_{k=1}^\infty a_k = L$$ and say, "the series $\sum_{k=1}^\infty a_k$ converges to $L$."
  If the sequence of partial sums diverges, we say that the series diverges.
• A series of the form $$\sum_{k=0}^\infty a_0 \, r^k$$ is called a geometric series.
• Suppose $a_0 \neq 0$. Then for a real number $r$ such that $|r| < 1$, the geometric series $$\sum_{k=0}^\infty a_0\, r^k$$ converges to $\frac{a_0}{1-r}$.
  For a real number $r$ where $|r| \geq 1$, the aforementioned geometric series diverges.
• Consider the series $$\sum_{k=0}^\infty a_k$$ and suppose $c$ is a nonzero constant. Then $$\sum_{k=0}^\infty a_k$$ and $$\sum_{k=0}^\infty c\,a_k$$ share a common fate: either both series converge, or both series diverge.
  Moreover, when $$\sum_{k=0}^\infty a_k$$ converges, $$\sum_{k=0}^\infty c \, a_k = c \cdot \sum_{k=0}^\infty a_k$$
• Suppose $$\sum_{k=0}^\infty a_k$$ and $$\sum_{k=0}^\infty b_k$$ are convergent series. Then $$\sum_{k=0}^\infty (a_k+b_k)$$ is convergent, and
  $$\sum_{k=0}^\infty (a_k+b_k)=\left( \sum_{k=0}^\infty a_k\right)+\left(\sum_{k=0}^\infty b_k\right)$$
• If $$\sum_{k=0}^\infty a_k$$ converges then $$\lim_{n\to\infty}a_n=0$$
• Consider the series $$\sum_{k=0}^\infty a_k$$ If the limit $$\lim_{n\to\infty}a_n$$ does not exist or has a value other than zero, then the series diverges.
  We'll usually call this theorem the "$n^{\text{th}}$ term test."
• The series $$\sum_{n=1}^\infty {1\over n} = \frac{1}{1} +\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots$$ is called the harmonic series.
• Consider the series $$\sum_{k=0}^\infty a_k$$ Assume the terms $a_k$ are non-negative. If the sequence of partial sums $s_n = a_0 + \cdots + a_n$ is bounded, then the series converges.
• Comparison Test
  Suppose that $a_n$ and $b_n$ are non-negative for all $n$ and that, for some $N$, whenever $n \geq N$, we have $a_n \leq b_n$.
  If $$\sum_{n=0}^\infty b_n$$ converges, so does $$\sum_{n=0}^\infty a_n$$ If $$\sum_{n=0}^\infty a_n$$ diverges, so does $$\sum_{n=0}^\infty b_n$$
• Cauchy Condensation Test
  Suppose $(a_n)$ is a non-increasing sequence of positive numbers. The series $$\sum_{n=1}^\infty a_n$$ converges if and only if the series $$\sum_{n=0}^\infty \left( 2^n a_{2^n} \right)$$ converges.
• $p$-series Test
  $$\sum_{n=1}^{\infty} {1\over n^p}\ \begin{cases}\mbox{converges} & \mbox{when $p>1$}\\ \mbox{diverges} & \mbox{when $p\leq1$} \end{cases}$$

Exercises 2.7

1. Explain why $$\sum_{n=1}^\infty {n^2\over 2n^2+1}$$ diverges.

Solution:

By $n^{\text{th}}$ test, $$\lim_{n\to\infty} {n^2\over 2n^2+1}={1\over2}\neq0$$
Therefore it diverges.

2. Explain why $$\sum_{n=1}^\infty {5\over 2^{1/n}+14}$$ diverges.

Solution:

By $n^{\text{th}}$ test, $$\lim_{n\to\infty} {5\over 2^{1\over n}+14}={5\over1+14}={1\over3}\neq0$$ Thus it diverges.

3. Explain why $$\sum_{n=1}^\infty {3\over n}$$ diverges.

Solution:

$$\sum_{n=1}^\infty {3\over n}=3\cdot\sum_{n=1}^{\infty}{1\over n}$$ which is a harmonic series.

4. Compute $$\sum_{n=0}^\infty {4\over (-3)^n}- {3\over 3^n}$$

Solution:

Geometric series:
$$\sum_{n=0}^\infty {4\over (-3)^n}- {3\over 3^n}$$
$$=\sum_{n=0}^{\infty} 4\cdot(-{1\over3})^n-3\cdot({1\over3})^n$$
$$=4\cdot{1\over 1-(-{1\over3})}-3\cdot{1\over 1-{1\over3}}$$
$$=4\times{3\over4}-3\times{3\over2}=-{3\over2}$$

5. Compute $$\sum_{n=0}^\infty {3\over 2^n}+ {4\over 5^n}$$

Solution:

Geometric series:
$$\sum_{n=0}^\infty {3\over 2^n}+ {4\over 5^n}$$
$$= 3\cdot{1\over 1-{1\over2}}+4\cdot{1\over 1-{1\over5}}=6+5=11$$

6. Compute $$\sum_{n=0}^\infty {4^{n+1}\over 5^n}$$

Solution:
Geometric series:
$$\sum_{n=0}^\infty {4^{n+1}\over 5^n}$$
$$=\sum_{n=0}^{\infty}4\cdot({4\over5})^n=4\times{1\over 1-{4\over5}}=20$$

7. Compute $$\sum_{n=0}^\infty {3^{n+1}\over 7^{n+1}}$$

Solution:

Geometric series:
$$\sum_{n=0}^\infty {3^{n+1}\over 7^{n+1}}$$
$$=\sum_{n=0}^\infty {3\over7}\cdot{3^{n}\over 7^{n}}={3\over7}\times{1\over 1-{3\over7}}={3\over4}$$

8. Compute $$\sum_{n=1}^\infty \left({3\over 5}\right)^n$$

Solution:

Geometric series:
$$\sum_{n=1}^\infty \left({3\over 5}\right)^n$$
$$=\sum_{n=0}^\infty \left({3\over 5}\right)^n-1$$
$$={1\over 1-{3\over5}}-1={3\over2}$$
Alternatively, $$\sum_{n=1}^\infty \left({3\over 5}\right)^n={{3\over5}\over 1-{3\over5}}={3\over2}$$

9. Compute $$\sum_{n=1}^\infty {3^n\over 5^{n+1}}$$

Solution:

Geometric series:
$$\sum_{n=1}^\infty {3^n\over 5^{n+1}}$$
$$=\sum_{n=0}^\infty {1\over5}\cdot{3^n\over 5^{n}}-{1\over5}$$
$$={1\over5}\times{1\over 1-{3\over5}}-{1\over5}={3\over10}$$
Alternatively, $$\sum_{n=1}^\infty {3^n\over 5^{n+1}}={1\over5}\times{{3\over5}\over 1-{3\over5}}={3\over10}$$

Additional Exercises

1. Evaluate $$\sum_{n=5}^{\infty}(-{4\over7})^n$$

Solution:

Geometric series:
$$\sum_{n=5}^{\infty}(-{4\over7})^n={(-{4\over7})^5\over 1-(-{4\over7})}=-{1024\over26411}$$

2. Test $$\sum_{n=2}^{\infty}-8\cdot({6\over11})^n$$

Solution:

$$\sum_{n=2}^{\infty}-8\cdot({6\over11})^n=-8\cdot\sum_{n=2}^{\infty}({6\over11})^n$$ which is a geometric series and $r < 1$, therefore it converges.

3. Evaluate $$\sum_{i=2}^{\infty}{12\over 9i^2+21i+10}$$

Solution:

$$\sum_{i=2}^{\infty}{12\over 9i^2+21i+10}=\sum_{i=2}^{\infty}{12\over (3i+5) (3i+2)}$$
$$=\sum_{i=2}^{\infty}4\cdot({1\over 3i+2}-{1\over 3i+5})=4\times{1\over8}={1\over2}$$

4. Test $$\sum_{m=3}^{\infty}{(7m+5)\cdot(m-8)\over(5m+4)\cdot(5m-7)}$$

Solution:

By $n^{\text{th}}$ test: $$\lim_{m\to\infty}{(7m+5)\cdot(m-8)\over(5m+4)\cdot(5m-7)}={7\over25}\neq0$$
Thus it diverges.

5. Test $$\sum_{n=0}^{\infty}{5\over 7n+42}$$

Solution:

$$\sum_{n=0}^{\infty}{5\over 7n+42}={5\over7}\cdot\sum_{n=0}^{\infty}{1\over n+6}$$ which is a harmonic series. Thus it diverges.

6. Test $$\sum_{n=5}^{\infty}{2(n^2+2)\over 7^n}$$

Solution:

Comparison test:
$$\sum_{n=5}^{\infty}{2(n^2+2)\over 7^n}=\sum_{n=5}^{\infty}{2n^2+4\over 7^n}\leq\sum_{n=5}^{\infty}{2^n\over 7^n},\ \text{when}\ n\geq7$$
And $$\sum_{n=5}^{\infty}{2^n\over 7^n}=\sum_{n=5}^{\infty}({2\over7})^n$$
is a geometric series which is convergent. Thus $$\sum_{n=5}^{\infty}{2(n^2+2)\over 7^n}$$ is convergent, too.